Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I prove that if $a$, $b$ are elements of group, then $o(ab) = o(ba)$?

For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of $ab$ is $2$ and do the steps that lead me to conclude that $ab=ba$, so the orders must be the same. Is that the right way to do it?

share|cite|improve this question
Why on earth are you assuming that the order of $ab$ is $2$? – Chris Eagle Dec 8 '12 at 11:14
This question is related to… – user26857 Apr 22 '13 at 21:33

Here's an approach that allows you to do some hand-waving and not do any calculations at all. $ab$ and $ba$ are conjugate: indeed, $ba=a^{-1}(ab)a$. It is obvious (and probably already known at this point) that conjugation is an automorphism of the group, and it is obvious that automorphisms preserve orders of elements.

share|cite|improve this answer
Calling this «hand-waving» is quite misguided! – Mariano Suárez-Alvarez Apr 22 at 7:33

Hint: Suppose $ab$ has order $n$, and consider $(ba)^{n+1}$.

Another hint is greyed out below (hover over with a mouse to display it):

Notice that $(ba)^{n+1} = b(ab)^na$.

share|cite|improve this answer

If $(ab)^n=e$ then $(ab)^na=a$. Since $(ab)^na=a(ba)^n$, $(ba)^n=e$. This proves that the order of $ba$ divides the order of $ab$. By symmetry, the order of $ab$ divides the order of $ab$. Hence the order of $ab$ and the order of $ba$ coincide.

share|cite|improve this answer
I think the OP should note that the orders of $a$ and $b$ are both finite. – Babak S. Nov 15 '12 at 20:58
@BabakSorouh Why? The order of ab may be finite while those of a and b are infinite. – Did Nov 15 '12 at 21:11
@Did. I would like to know of a specific example of a group with elements a and b with infinite order but the order of ab is finite. I am not disputing what you are stating. I am a beginning student of algebra and I need " a good stock of examples". I can see that if a and b are inverses of each other then ab=e, but I was hoping for an example where a does not equal b inverse. – Geoffrey Critzer Jun 5 '15 at 10:28
@GeoffreyCritzer Try $b=a^{-1}$. – Did Jun 5 '15 at 10:33
@GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)\mapsto(y,x)$. – Did Jun 5 '15 at 10:44

By associativity, $(ab)^p=a(ba)^{p-1}b$ for $p\geqslant 1$. If $(ab)^p=e$ then $a(ba)^{p-1}b=e$, so $a(ba)^p=a$ and $(ba)^p=e$. We conclude that for $p\geqslant 1$, $$(ab)^p=e\Leftrightarrow (ba)^p=e.$$

share|cite|improve this answer

protected by Zev Chonoles Apr 22 at 6:54

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.